To 1.0 L of a 0.38 M solution of HClO2 is added 0.18 mol of NaClO. Calculate the [HClO2] at equilibrium.
If the salt is NaClO2 I would use the Henderson-Hasselbalch equation for buffer solutions. Try that and compare with the ICE method.
Just to make sure, is that NaClO or NaClO2?
If the 0.18 mole is NaClO2, I get 0.25M HClO2 at equilibrium.
Note => In the process of calculating the HClO2 concentration using the ICE table analysis, Ka = 1.1E-2 which indicates by the simplification rule, one can not drop the 'x' in the equilibrium row when calculating the Hydronium ion concentration. Give a try... Post your work if ya get stuck.
One must 'assume' a pH value with pKa from Ka = 1.1E-2. I used pH = 1.8 (pH = 1.5 - 2.0 is typical of chloric acid solutions in the 0.1 - 0.2M range) & pKa = -log(Ka) = -log(1.1E-2) = 1.96 and got [HOCl2] = 0.25M. Same as the ICE table analysis. Great suggestion DrBob!
Well, let me see if I got this straight. We have some HClO2 and we're adding some NaClO to the mix. It sounds like the start of a pretty interesting party! 🎉
But I guess I should get serious about this chemistry stuff. To calculate the concentration at equilibrium, we need to use the stoichiometry of the reaction.
The balanced equation for the reaction between HClO2 and NaClO is:
HClO2 + NaClO → NaCl + HClO3
According to the equation, 1 molecule of HClO2 reacts with 1 molecule of NaClO, resulting in the formation of 1 molecule of HClO3.
So, if we start with 0.18 mol of NaClO, we know that 0.18 mol of HClO2 will react with it as well.
We started with 1.0 L of a 0.38 M solution of HClO2, which means we have 0.38 moles of HClO2 in that 1.0 L.
Since 0.18 mol of HClO2 reacts with 0.18 mol of NaClO, the concentration of HClO2 at equilibrium will be 0.38 - 0.18 = 0.20 M.
So, the final [HClO2] at equilibrium is 0.20 M.
Hope that helps! And remember, next time you throw a party, don't forget the stoichiometry! 🎉😄
To calculate the concentration of HClO2 at equilibrium, we need to determine the changes in the concentrations of HClO2 and NaClO after the reaction.
The balanced chemical equation for the reaction between HClO2 and NaClO is:
HClO2 + NaClO -> ClO2^- + Na^+ + H2O
From the equation, we can see that for every 1 mole of HClO2, 1 mole of ClO2^- is produced. Therefore, the number of moles of ClO2^- formed is equal to the number of moles of HClO2 consumed.
Given:
Initial volume of HClO2 solution = 1.0 L
Initial concentration of HClO2 = 0.38 M
Number of moles of HClO2 = initial concentration x volume = 0.38 mol/L x 1.0 L = 0.38 mol
0.18 mol of NaClO is added. Since the reaction is stoichiometric, the number of moles of ClO2^- produced is the same as the number of moles of HClO2 consumed. Therefore, the total number of moles of ClO2^- formed is 0.18 mol.
At equilibrium, the total volume of the solution will be the initial volume of the HClO2 solution plus the volume of NaClO added. Since the volume of NaClO is not given, we need additional information to determine the final total volume of the solution.
Once the final total volume is determined, we can calculate the concentration of HClO2 at equilibrium by dividing the number of moles of HClO2 remaining by the final total volume.
Please provide the volume of NaClO added to the solution to proceed with the calculation.